Câu hỏi của Shinichi Kudo

Question

Cho ba số a, b, c thỏa mãn điều kiện: a+b+c=0;

a2 + b2 + c2 = 14

Tính P = 1 + a4 + b4 + c4

Hàn Thiên Vũ · Accepted Answer

Có:

$a+b+c=0$

$\Rightarrow\left(a+b+c\right)^2=0$

$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)$

$\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2$

$\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left[a^2b^2+b^2c^2+a^2c^2+abc\left(a+b+c\right)\right]$

$\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)$

$\Rightarrow a^4+b^4+c^4+1=2\left(a^2b^2+b^2c^2+a^2c^2\right)+1$

Có:

$\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2$

$\Rightarrow4\left(ab+bc+ac\right)^2=196$

$\Rightarrow\left(ab+bc+ac\right)^2=49$

$\Rightarrow a^2b^2+b^2c^2+a^2c^2=49$

$\Rightarrow a^4+b^4+c^4+1=2\left(a^2b^2+b^2c^2+a^2c^2\right)+1$

$\Rightarrow a^4+b^4+c^4+1=2.49+1$

$\Rightarrow a^4+b^4+c^4+1=99$Câu trả lời: Có:

$a+b+c=0$

$\Rightarrow\left(a+b+c\right)^2=0$

$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)$

$\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2$

$\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left[a^2b^2+b^2c^2+a^2c^2+abc\left(a+b+c\right)\right]$

$\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)$

$\Rightarrow a^4+b^4+c^4+1=2\left(a^2b^2+b^2c^2+a^2c^2\right)+1$

Có:

$\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2$

$\Rightarrow4\left(ab+bc+ac\right)^2=196$

$\Rightarrow\left(ab+bc+ac\right)^2=49$

$\Rightarrow a^2b^2+b^2c^2+a^2c^2=49$

$\Rightarrow a^4+b^4+c^4+1=2\left(a^2b^2+b^2c^2+a^2c^2\right)+1$

$\Rightarrow a^4+b^4+c^4+1=2.49+1$

$\Rightarrow a^4+b^4+c^4+1=99$

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